Answer :
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Explanation:-
Given that \( a + b + c = 0 \), we need to show that \( (a+b-c)^3 + (b+c-a)^3 + (c+a-b)^3 = -24abc \).
To start, notice that if \( a + b + c = 0 \), then \( c = -(a+b) \).
Substitute \( c = -(a+b) \) into \( (a+b-c)^3 + (b+c-a)^3 + (c+a-b)^3 \):
\[ (a+b-c)^3 = (a+b-(-(a+b)))^3 = (2(a+b))^3 = 8(a+b)^3, \]
\[ (b+c-a)^3 = (b+(-(a+b))-a)^3 = (2(b+c))^3 = 8(b+c)^3, \]
\[ (c+a-b)^3 = ((-(a+b))+a-b)^3 = (2(c+a))^3 = 8(c+a)^3. \]
Now add these:
\[ (a+b-c)^3 + (b+c-a)^3 + (c+a-b)^3 = 8(a+b)^3 + 8(b+c)^3 + 8(c+a)^3. \]
Since \( a+b+c = 0 \), we have \( c = -(a+b) \). Substituting this into the sums:
\[ (a+b)^3 + (b+c)^3 + (c+a)^3 = a^3 + b^3 + c^3. \]
Since \( c = -(a+b) \), \( a^3 + b^3 + c^3 = 3abc \).
Therefore,
\[ 8(a^3 + b^3 + c^3) = 8 \cdot 3abc = 24abc. \]
So, \( (a+b-c)^3 + (b+c-a)^3 + (c+a-b)^3 = -24abc \).
This completes the proof.
Step-by-step explanation:
do this as shown above,you will find your answer.
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